extract function with variable arguments

Question

Given I have the function:

void A::IAmCool(int x, ...)
{
        va_list args; 
        va_start (args, x);   
        va_end (args);
}

How do I pass var-args from one function to another? I'm looking for something like this:

void A::extractedFunction() /* this doesn't work */
{
        va_list args; 
        va_start (args, ?????);   
        va_end (args);
}

void A::IAmCool(int x, ...)
{
        extractedFunction();
}

Is this even possible? I have tried making the function inline but that doesn't work.

@djechlin: I'm going to guess is referring to the Eclipse-style refactoring facility. — Oliver Charlesworth
– Oliver Charlesworth, Commented May 25, 2012 at 20:53
If you are using C++, do not write variadic functions. End of story. Variadic functions completely lack type safety. There are other patterns (e.g., what the IO streams or Boost Format do) that are far superior. [If you have a compiler with variadic templates support, then that's a different matter altogether, but you've said that you're using Visual C++, which does not support variadic templates.] — James McNellis
– James McNellis, Commented May 25, 2012 at 21:03

Kerrek SB · Accepted Answer · 2012-05-25 21:12:41Z

1

The usual pattern is to implement your main workhorse function taking a valist, and the variadic function only as "decoration". Then you can use the main function directly from third-party call sites. Example:

#include <cstdarg>

int vgizmo(int a, std::va_list ap)
{
    // main implementation here!
}

int gizmo(int a, ...)  // interface function
{
    std::va_list ap;
    va_start(ap, a);
    int r = vgizmo(a, ap);
    va_end(ap);
    return r;
}

void some_other_stuff(bool q, char const * fmt, ...)
{
    std::va_list ap;
    va_start(ap, fmt);

    // ...

    int b = vgizmo(x, ap);   // third parties use vgizmo directly

    // ...

    va_end(ap);
}

answered May 25, 2012 at 21:12

Kerrek SB

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2 Comments

Brian McFarland Over a year ago

See vprintf() for a widespread example of this approach... on linux, just man vprintf. And one important thing to note is that in spite of looking like a pass-by-value function call, va_arg() is a macro that modifies the va_list. In other words, you can't pass the same va_list into two different functions without a va_end() and another va_start() in between. Also, the linux man page man stdarg is pretty good.

Kerrek SB Over a year ago

@BrianMcFarland: You can use va_copy to create multiple copies of a given valist.

user529758 · Accepted Answer · 2012-05-25 21:09:04Z

In C-like languages (C, C++, Objective-C etc.) you can't directly pass variadic arguments. However, you may (and if you're developing an external library, or you just want to facilitate the work of reverse engineers you should) supply a non-variadic version of your function, and make the variadic one just wrap it, so that you can pass down as many arguments as needed to your function -- using va_list. Example:

// variadic function
void variadic_func(int nargs, ...)
{
    // just wrap the non-variadic one
    va_list args;
    va_start(args, nargs);
    non_variadic_func(nargs, args);
    va_end(args);
}

// non-variadic function
void non_variadic_func(int nargs, va_list args)
{
    // do what you want with `args'
}

// you can pass down varargs like this:
void outmost_caller_func(int nargs, ...)
{
    // since you can't pass down the `...', you create a va_list argument list
    va_list args;
    va_start(args, nargs);
    // and call the non-variadic version of your function, just like the wrapper
    // would do (anyway, the wrapper is only for convenience...)
    non_variadic_func(nargs, args);
    va_end(args);
}

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